Question

2^(x-1) + 3/8×2^(2x+1)=784

Na minha resposta, deu x=5... na resolução de vocês, também dá esse resultado? Efetuem as operações, por favor.
Obrigado!.

Answer 1

Olá Bruno.


Organizando e resolvendo a equação exponencial:

$\mathsf{2^{x-1}+\dfrac{3}{8}\cdot 2^{2x+1}=784}\\\\\mathsf{\underbrace{\mathsf{2^x}}\cdot 2^{-1}+\dfrac{3}{8}\cdot 2^x\cdot2^x\cdot2^1=784}\\\mathsf{~~y}\\\\\mathsf{y\cdot \dfrac{1}{2}+\dfrac{3}{\diagup\!\!\!\!8}\cdot y\cdot y\cdot\diagup\!\!\!\!2=784}\\\\\mathsf{\dfrac{y}{2}+\dfrac{3}{4}\cdot y^2=784}\\\\\mathsf{\dfrac{2y+3y^2}{\diagdown\!\!\!\!4}=\dfrac{3.136}{\diagdown\!\!\!\!4}}\\\\\\\mathsf{3y^2+2y=3.136}\\\\\mathsf{3y^2+2y-3.136}$

$\mathsf{\Delta=(2)^2-4\cdot3\cdot(-3.136)}\\\mathsf{\Delta=4+37.632}\\\mathsf{\Delta=37.636}\\\\\\\mathsf{y=\dfrac{-2\pm\sqrt{37.636}}{2\cdot3}}\\\\\\\mathsf{y^+=\dfrac{-2+194}{6}\qquad\qquad\qquad\qquad y^-=\dfrac{-2-194}{6}}\\\\\\\mathsf{y^+=\dfrac{192}{6}\qquad\qquad\qquad\qquad\qquad y^-=\dfrac{-196}{6}}\\\\\\\mathsf{y^+=32\qquad\qquad\qquad\qquad\qquad~~ y^-=-\dfrac{98}{3}}$


$\mathsf{2^x=y^+}\\\\\mathsf{2^x=32}\\\\\mathsf{2^x=2^5}\\\\\boxed{\mathsf{x=5}}\\\\\\\\\mathsf{2^x=-\dfrac{98}{3}~~\gets~~absurdo!}\\\\\\\\\boxed{\mathsf{S:\{5\}}}$


Dúvidas? comente.